(D) According to the Arrhenius equation,$K_1 = A_1 e^{-Ea_1/RT}$ and $K_2 = A_2 e^{-Ea_2/RT}$.Dividing the two equations: $\frac{K_1}{K_2} = \frac{A_1

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(D) According to the Arrhenius equation,$K_1 = A_1 e^{-Ea_1/RT}$ and $K_2 = A_2 e^{-Ea_2/RT}$.Dividing the two equations: $\frac{K_1}{K_2} = \frac{A_1}{A_2} e^{(Ea_2 - Ea_1)/RT}$.Given $Ea_2 = 2 Ea_1$,substituting this into the exponent gives: $\frac{K_1}{K_2} = \frac{A_1}{A_2} e^{(2 Ea_1 - Ea_1)/RT} = \frac{A_1}{A_2} e^{Ea_1/RT}$.Therefore,$K_1 = A' K_2 e^{Ea_1/RT}$,where $A' = \frac{A_1}{A_2}$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

Easy

Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

A
$K_1 = 2 K_2 e^{Ea_2/RT}$
B
$K_2 = K_1 e^{Ea_1/RT}$
C
$K_2 = K_1 e^{Ea_2/RT}$
D
$K_1 = A' K_2 e^{Ea_1/RT}$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

If the rate of a reaction doubles when the temperature increases from $298 \, K$ to $308 \, K$,the activation energy of the reaction is ........... $kJ \, mol^{-1}$.

Medium

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For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

Medium

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The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :
($R$ is gas constant)

MediumJEE MAIN 2020

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Which one of the following is true for an exothermic reaction $A \rightleftharpoons B$,if $E_f$ and $E_b$ are the activation energies of forward and backward reactions respectively?

EasyTS EAMCET 2011

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For the reaction $2NO + Cl_2 \rightarrow 2NOCl$,the rate is given by $Rate = K[NO]^2[Cl_2]$. How can the rate constant $(K)$ of the reaction be increased?

Easy

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Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:$A \xrightarrow{K_1} B$,activation energy: $Ea_1$$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

Similar Questions

If the rate of a reaction doubles when the temperature increases from $298 \, K$ to $308 \, K$,the activation energy of the reaction is ........... $kJ \, mol^{-1}$.

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :($R$ is gas constant)

Which one of the following is true for an exothermic reaction $A \rightleftharpoons B$,if $E_f$ and $E_b$ are the activation energies of forward and backward reactions respectively?

For the reaction $2NO + Cl_2 \rightarrow 2NOCl$,the rate is given by $Rate = K[NO]^2[Cl_2]$. How can the rate constant $(K)$ of the reaction be increased?

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Reactant $(A)$ produces two products. If $Ea_2 = 2 Ea_1$,then $K_1$ and $K_2$ are related as:
$A \xrightarrow{K_1} B$,activation energy: $Ea_1$
$A \xrightarrow{K_2} C$,activation energy: $Ea_2$

The rate constant $(k)$ of a reaction is measured at different temperatures $(T),$ and the data are plotted in the given figure. The activation energy of the reaction in $kJ\, mol^{-1}$ is :
($R$ is gas constant)